Solution: The focus of parabola is $(4,0)$. Let slope of focal chord be $m$. Equation of focal chord is $y=m(x-4)$. It is tangent to the circle then

$\begin{array}{rl}& \left|\frac{6m-4m}{\sqrt{m^2+1}}\right|=\sqrt{2}\\ & 4m^2=2(m^2+1)\\ & 2m^2=2\\ & m=\pm 1\end{array}$

Answer: a

Solution: $\mathrm{ax}=1+\mathrm{by}-2\sqrt{\mathrm{by}}$

$\begin{array}{rl}& (ax-by-1{)}^2=(-2\sqrt{by}{)}^2\\ & a^2{x}^2+b^2{y}^2+1-2abxy-2ax+2by=4by\\ & a^2{x}^2-2abxy+b^2{y}^2-2ax-2by+1=0\\ & \begin{array}{r}\mathrm{\Delta }=\left|\begin{array}{ccc}{a}^2& -ab& -a\\ -ab& b^2& -b\\ -a& -b& 1\end{array}\right|\\ & =a^2(b^2-b^2)+ab(-ab-ab)-a(a{b}^2+a{b}^2)\\ & =0-2a^2{b}^2-2a^2{b}^2\\ & =-4a^2{b}^2\ne 0\end{array}\\ & h^2-ab=(-ab{)}^2-\left(a^2\right)\left(b^2\right)=0\\ & \therefore \text{ It is a parabola }\end{array}$

Answer: b

Solution: ${\mathrm{y}}^2+4\mathrm{y}=-4\mathrm{x}-2$

$(y+2{)}^2=-4x-\frac{1}{2}$

$y^2=-4AX$

Equation of directrix is $\mathrm{X}=$ Ai.e. $\mathrm{x}-\frac{1}{2}=1$

$2\mathrm{x}-3=0$ or $\mathrm{x}=\frac{3}{2}$

Answer: d

Solution: Let $\mathrm{P}({\mathrm{at}}^2,2\mathrm{at})$ lies on the parabola

${\mathrm{y}}^2=4\mathrm{ax}$

Mid point of PS is Q.

$\mathrm{h}=\frac{\mathrm{at}{}^2+\mathrm{a}}{2},\mathrm{k}=\frac{0+2\mathrm{at}}{2}$

$\frac{2\mathrm{h}-\mathrm{a}}{\mathrm{a}}={\mathrm{t}}^2,\frac{\mathrm{k}}{\mathrm{a}}=\mathrm{t}$

$\frac{2\mathrm{h}-\mathrm{a}}{\mathrm{a}}={\frac{\mathrm{k}}{\mathrm{a}}}^2$

$\mathrm{a}(2\mathrm{h}-\mathrm{a})={\mathrm{k}}^2$

Locus of $(h,k)$ is $y^2=2ax-\frac{a}{2}$

Equaiton of directrix is $x-\frac{a}{2}=-\frac{a}{2}$

$x=0$

Answer: c

Solution: Equation of tangent of the parabola $y^2=4x$ is

$\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$

This equation passes through $(1,4)$ i.e.

$4=\mathrm{m}+\frac{1}{\mathrm{m}}$

$m^2-4m+1=0$

${\mathrm{m}}_1\cdot {\mathrm{m}}_2=1$ and ${\mathrm{m}}_1+{\mathrm{m}}_2=4$

Angle between the two tangents is $\tan \theta =\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

$\tan \theta =\left|\frac{\sqrt{(4{)}^2-4}}{1+1}\right|=\frac{\sqrt{12}}{2}=\sqrt{3}$

$\therefore \theta =\frac{\pi }{3}$

Answer: c

Solution: Equation of tangent at $(1,7)$ to the curve $y=x^2+6$ is

$\frac{\mathrm{y}+7}{2}=\mathrm{x}+6$

$2\mathrm{x}-\mathrm{y}+5=0$

This line also touches the circle i.e.

Equation of normal of circle passing through

centre $(-8,-6)$.

$\mathrm{x}+2\mathrm{y}+\lambda =0$

$-8-12+\lambda =0$

$\lambda =20$

$\therefore \mathrm{x}+2\mathrm{y}+20=0$

$\mathrm{Q}$ is intersection point of (1) and (2)

$\mathrm{x}=-6$,

$y=-7$

$\mathrm{Q}(-6,-7)$

Answer: d

Solution: Let eq ${\mathrm{q}}^{\mathrm{n}}$ of tangent of parabola be $\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$ is also a tangent to the circle then $\left|\frac{3m+\frac{1}{m}}{\sqrt{1+m^2}}\right|=2\sqrt{2}$ $\frac{{(3m^2+1)}^2}{m^2}=8(1+m^2)$

${\mathrm{m}}^4-2{\mathrm{m}}^2+1=0$

${\mathrm{m}}^2=1\mathrm{m}=\pm 1$ Two common tangents are possible.

Solution: We know that $2\mathrm{a}=\frac{2\mathrm{xSAxSB}}{\mathrm{SA}+\mathrm{SB}}$

$2a=\frac{2bc}{b+c}\Rightarrow \begin{array}{rl}ab& =bc-ac\\ ab& =(b-o)c\end{array}$

$\mathrm{ab}+\mathrm{ac}=\mathrm{bc}$

$\frac{ab}{b-a}=c$

Answer: d

Solution: Let $y=mx+\frac{2}{m}$ be tangent to $y^2=8x$. Since circle intersects the parabola orthogonally. So this tangent is the normal for the circle. Every normal of the circle passes through its centre. So centre $(1,3)$.

$\begin{array}{rl}& 3=m+\frac{2}{m}{m}^2-3m+2=0\\ & (m-2)(m-1)=0\\ & m=1,2\\ & y=x+2\text{ or }y=2x+1\end{array}$

Answer: a